Update: The negative answer to the following question has been provided by Matthew Daws, who won, but also rejected, the bounty of 100 euro that I set over the question.

Let $\mathcal M(\mathbb Z)$ be the set of all finitely additive probability measures on the power set of $\mathbb Z$. Let $\phi:\mathbb Z\rightarrow\mathbb R$ be nonnegative and bounded. Observe that $\phi$ is integrable with respect to any $\mu\in\mathcal M(\mathbb Z)$. Let me say that $\mu$ is $\phi$-translation invariant if for all $y\in\mathbb Z$ one has

A simple observation which might be silly because I don't know if Fubini's theorem holds for finitely additive probability measures: If we are in a situation where we can use Fubini's theorem the mapping in the question is constant since $$\int\int\phi(x+y)d\nu(x)d\mu(y) = \int\int\phi(x+y)d\mu(y)d\nu(x)$$ $$= \int\int\phi(y)d\mu(y)d\nu(x) = \int\phi(y)d\mu(y).$$
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Tapio RajalaSep 3 '11 at 8:34

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Unfortunately Fubini's theorem does not hold. For intance, let $\phi=\chi_{\mathbb N}, $\mu$ a invariant measure such that $\int\phi d\mu=1$ e $\vu$ an invariant measure such that $\int\phi d\vu=0$. The lackness of Fubini's theorem is, at the end, the point.
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Valerio CapraroSep 3 '11 at 8:44

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I think the title should be "Do invariant measures maximize the integral? (Bounty offered)" . That phrasing is slightly less crass, and stirs more curiosity; people will read through the post to find out what the bounty is and how much; I think the present phrasing of the title is in the grey area of acceptability on MathOverflow. Gerhard "It May Be Just Me" Paseman, 2011.11.18
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Gerhard PasemanNov 18 '11 at 17:27

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Indeed he did Alain. And how many of those were phrased in a fashion appropriate for MathOverflow? I am not opposing the occasional practice; I just think the presentation should be improved. Gerhard "It Might Be Others Too" Paseman, 2011.11.18
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Gerhard PasemanNov 18 '11 at 18:58

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There is a meta discussion, specifically discussing the bounty, at tea.mathoverflow.net/discussion/1212/…. My suggestion would be to explicitly spell out that the bounty is entirely independent of the usual mathematical conventions regarding acknowledgement and priority.
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Scott Morrison♦Nov 18 '11 at 21:28

So if $\phi\in A^*$ if positive then $\mu\in I_\phi$ if and only if $\mu\cdot\phi
= \mu(\phi) 1$. This follows, as writing $\delta_x\in A=\ell^1(\mathbb Z)$ for the point mass at $x\in\mathbb Z$, we have
\[ (\phi\cdot\delta_x)(\delta_y) = \phi(\delta_{x+y}) \implies
(\mu\cdot\phi)(\delta_x) = \mu(\phi\cdot\delta_x)
= \int \phi(x+y) \ d\mu(y). \]
So the condition that $\mu\in I_\phi$ becomes that $(\mu\cdot\phi)(\delta_x)$ is constant in $x$, which is seen to be equivalent to $\mu\cdot\phi = \mu(\phi) 1$.

Similarly, your map $F$ is just $F(\nu) = (\mu\Box\nu)(\phi)$.

As you allude to, it's known that $\lambda\Box\mu \not= \mu\Box\lambda$ for arbitrary $\lambda,\mu$. However, we say that $f\in A^*$ is "weakly almost periodic" (WAP) if $(\lambda\Box\mu)(f) = (\mu\Box\lambda)(f)$ for all $\mu,\lambda\in A^{**}$. So if $\phi$ is WAP and $\mu\in I_\phi$ then for any $\nu\in M(\mathbb Z)$,
\[ F(\nu) = (\mu\Box\nu)(\phi) = (\nu\Box\mu)(\phi) = \nu(\mu\cdot\phi)
= \nu(1) \mu(\phi) = \mu(\phi), \]
as $\nu$ is a probability measure. So actually $F$ is constant on $M(\mathbb Z)$ and so certainly attains its maximum at a point of $I_\phi$.

So, to be interesting, we need to ask the question for $\phi$ which are not WAP. An alternative characterisation of $\phi$ being in WAP is that the set of translates of $\phi$ in $\ell^\infty(\mathbb Z)$ forms a relatively weakly compact set. A nice characterisation of Grothendieck shows that this is equivalent to
\[ \lim_n \lim_m \phi(x_n+y_m) = \lim_m \lim_n \phi(x_n+y_m) \]
whenever all the limits exist, for sequences $(x_n),(y_m)$ in $\mathbb Z$. If $\phi$ is the indicator function of $\mathbb N$, then it's not in WAP.

We may as well assume that $\|\phi\|_\infty=1$.
Another "easy" case is when we can find $\nu\in I_\phi$ with $\nu(\phi)=1$. Then $F(\nu) = \mu(\nu\cdot\phi) = \mu(1) \nu(\phi) = 1$; while for any $\lambda\in M(\mathbb Z)$, clearly $|F(\lambda)| = |\mu(\lambda\cdot\phi)| \leq 1$ as $\mu$ is a probability measure, and $\lambda\cdot\phi$ is bounded by $1$ (again, as $\lambda$ is a probability measure and $\phi$ is bounded by $1$). Notice that this case covers your example of when $\phi$ is the indicator function of $\mathbb N$.

So a test case is to find $\phi$ not in WAP and with $\nu(\phi)<\|\phi\|_\infty$ for all $\nu\in I_\phi$ (notice that $I_\phi$ is always non-empty, as $\mathbb Z$ is amenable). Do you have an example of such a $\phi$?

Actually, if $\phi$ is the indicator function of the even natural numbers, then that's an example. And that leads to my (hopeful) counter-example.

Matthew, I have just seen your answer. Give me some time to go through the details. Let me understand well: are you claiming that the answer is negative and give a counterexample in the Edit?
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Valerio CapraroNov 18 '11 at 19:10

Yes. Actually, let me make a final edit-- I'll make the counter-example clearer (I'll put it into your notation) and make a revised conjecture...
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Matthew DawsNov 18 '11 at 20:30

Hi Matthew, sorry for the delayed answer but I was mainly away this weekend. I am now sure that your counter-example is good. Please, contact me privately for the bounty. I am thinking about the missing property.. indeed, for my application, I have some stronger property and so there might be still a positive answer.. let me think about for a while.
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Valerio CapraroNov 20 '11 at 17:31

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@Valerio: I'm glad the counter-example seems okay. I must say that I don't think I have done anything like enough work here to justify taking 100 euros off you (I hope that doesn't seem churlish). But what I will accept is, if you are ever in the North of England (or we meet at a conference), then you can buy me a drink or two... (and let us hope that still costs less than 100 euros after the current financial mess!) I'll email you shortly...
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Matthew DawsNov 20 '11 at 20:07